Find area of triangle given one angle and the lengths its altitude divides the opposite side into

Question

In triangle $\triangle ADC$, $DB$ is perpendicular to $AC$ at $B$ so that $AB=2$ and $BC=3$ as shown in the figure. Furthermore, $\angle ADC=45^\circ$ . Find the area of $\Delta ADC$.

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Tried the following things:

1. Drawing medians and altitudes of the right triangles formed.
2. Apollonius's theorem.
3. Stewart's theorem.
4. Sine rule.
5. Cosine rule.
6. Noticing that $3+2$ equals to $5$, I extended segment $BD$ to form a 3-4-5 triangle and try to proceed.

Spent around 1.5 hours on it. In each of the cases, I wasn't able to proceed after a certain point.

Answer 1

$1=\tan(\alpha + \beta) = {\tan(\alpha)+\tan(\beta)\over 1-\tan(\alpha)\tan(\beta)}$

${\tan(\alpha) \over \tan(\beta)} = {2\over 3}$

Solve them you get $\tan(\beta)={1\over 2}$ so the height is $6$

Answer 2

In a triangle $\,\triangle ADC\,,\,$ $DB\,$ is perpendicular to $\,AC\,$ at $B\,$ so that $\,AB=2\,$ and $\,BC=3\,$ as shown in the figure. Furthermore, $\,\angle ADC=45\unicode{176}\,.$
Find the area of $\,\triangle ADC\,.$

Here is a solution which does not use trigonometric functions.

First of all we consider the circumscribed circle of $\,\triangle ADC\,.$

Since a central angle of a circle is twice any inscribed angle subtended by the same arc, it follows that

$\angle AOC=2\angle ADC=2\!\cdot45\unicode{176}=90\unicode{176}.$

So $\,\triangle AOC\,$ is a rectangle and isosceles triangle, hence it is a half-square triangle and consequently

$OA=OC=\dfrac{AC}{\sqrt2}=\dfrac5{\sqrt2}\,.\quad\left(\implies OD=\dfrac5{\sqrt2}\right)$

Moreover, $\,\triangle OCH\cong\triangle AOH\,$ are two congruent half-square triangles, therefore it results that

$CH=OH=HA=\dfrac{AC}2=\dfrac52\,.$

On the other hand, since $\,OHBK\,$ is a rectangle, we get that

$KB=OH=\dfrac52\;\;,$

$OK=HB=BC-CH=3-\dfrac52=\dfrac12\;\;,$

and, by applying Pythagoras’ theorem, it results that

$DK=\sqrt{OD^2-OK^2}=\sqrt{\left(\dfrac5{\sqrt2}\right)^2-\left(\dfrac12\right)^2}=\dfrac72\;.$

Therefore ,

$DB=DK+KB=\dfrac72+\dfrac52=6\;\;,$

and the area of $\,\triangle ADB\,$ is equal to

$\mathrm{Area}=\dfrac{AC\cdot DB}2=\dfrac{5\cdot6}2=15\;.$

Answer 3

With the cosine law:

Set $h = BD$. We have

$$\frac{\sqrt{2}}{2}=\cos(45^o) = \frac{DC^2+AD^2-CA^2}{2\ DC\ AD} = \frac{3^2+h^2+2^2+h^2-5^2}{2\sqrt{3^2+h^2}\sqrt{2^2+h^2}}$$

and solve for $h$ to get $h=6$. (One has to be careful: squaring the equations introduces an extraneous solution, which one needs to discard).

Answer 4

Rotate the triangle and draw it on the complex plane, so $z_A=h+2i$, $z_B=h$, $z_C=h-3i$, and $z_D=0$, where $h > 0$ is a real number representing the length of $BD$.

I will work in radian measure so $45°$ is $\pi/4$ radians. This is the difference in argument between $z_A$ and $z_D$, so we have

\begin{align} \arg(h+2i) - \arg(h-3i) &= \frac{\pi}{4} \\ \arg \left( \frac{h+2i}{h-3i} \right) &= \frac{\pi}{4} \tag{1} \\ \arg \left( \frac{(h+2i)(h+3i)}{(h-3i)(h+3i)} \right) &= \frac{\pi}{4} \tag{2} \\ \arg \left( \frac{(h^2-6) + (5h)i}{h^2 + 9} \right) &= \frac{\pi}{4} \tag{3} \\ \arg ((h^2-6) + (5h)i) &= \frac{\pi}{4} \tag{4} \\ \frac{5h}{h^2-6} &= \tan \frac{\pi}{4} = 1 \tag{5} \\ h^2 - 6 &= 5h \\ h^2 - 5h - 6 &= 0 \\ (h-6)(h+1) &= 0 \\ \end{align}

Since we want $h>0$, we must take $h=6$, so the area of $\triangle ADC$ is $\frac{1}{2}\cdot (2+3) \cdot 6=15$.

Steps: $(1)$ argument of quotient of complex numbers is the difference in their arguments (at least modulo $2 \pi$), $(2)$-$(3)$ multiply numerator and denominator by complex conjugate of denominator to make the denominator real and positive, $(4)$ argument of a complex number is unchanged by multiplying or dividing it by a real, positive number since you stay on the same ray from the origin, $(5)$ note that if $x+yi$ has argument between $-\pi/2$ and $\pi/2$ then $x>0$ and the formula for the argument is simply $\arctan(y/x)$. Life is more complicated elsewhere. Without using any trigonometric functions, we could have noted that $\arg(x+yi)=\pi/4$ is only true for the half-line $y=x$, $x>0$, and deduced immediately that $h^2-6=5h$.

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The above is basically equivalent to the following trigonometric argument using the arctangent addition formula:

$$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$

Note that this formula holds when $uv < 1$. Here if $u = \tan \angle ADB = 2/h$ and $v = \tan \angle CDB = 3/h$ then since both angles are less than $45°$, their tangents $u$ and $v$ will be less than one, and so will the product $uv$.

We have \begin{align} \arctan(2/h) + \arctan(3/h) &= 45° \\ \arctan\left( \frac{2/h + 3/h}{1 - (2/h)(3/h)} \right) &= 45° \\ \arctan\left( \frac{5/h}{1 - 6/h^2} \right) &= 45° \\ \arctan\left( \frac{5h}{h^2 - 6} \right) &= 45° \\ \end{align}

as before.